1. Fix a compact topological space $X$. Whether $\{(Y,\phi): Y \text{ is a compact metric space and }\phi:X\to Y \text{ is surjective continuous map}\}$ is a set?

  2. If $I$ is a set and for each $\alpha\in I$, $A_\alpha$ is a set. Is the collection $\{A_\alpha:\alpha\in I\}$ a set?

I am reading a proof of RRT for positive linear functionals articulated by V.S.Sunder and there the first collection was assumed as a set. I couldn't able to prove it. Nonetheless, I have modified the required set as $\mathcal{B}:=\{E\subset X: \exists \text{ cpt m.s. }Y, ~\pi:X\to Y \text{ is surjective, and }E=\pi^{-1}(A),~A\in\mathcal{B}_\pi\}$ and enjoyed the nice proof! of prof.V.S.Sunder.

Ref: V.S. Sunder, The Riesz representation theorem, Indian J. of Pure and Appl. Math., 39(6), December 2008, 467-481.

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    $\begingroup$ math.stackexchange.com/q/327533/156252 $\endgroup$ Nov 26, 2020 at 6:39
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    $\begingroup$ 2. Yes, this is the replacement axiom of ZF. For 1., literally no, though if homeomorphic spaces are regarded equal, then yes. $\endgroup$
    – Berci
    Nov 26, 2020 at 6:43
  • $\begingroup$ @Berci, Thank you... $\endgroup$ Nov 26, 2020 at 6:59
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    $\begingroup$ The Hilbert Cube is $H=[0,1]^{\Bbb N}$ with the usual topology on $[0,1$] and the Tychonoff product on $ H$..... $H$ is compact & metrizable. Any 2nd-countable $T_{3\frac {1}{2}}$ space is homeomorphic to a subspace of $H$. In particular a space is compact & metrizable iff it is homeomorphic to a closed subspace of $H.$ $\endgroup$ Nov 26, 2020 at 10:48
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    $\begingroup$ See also math.stackexchange.com/questions/3470502/…. $\endgroup$
    – Paul Frost
    Nov 27, 2020 at 10:51

1 Answer 1


Your question reaches down to the foundation of set theory. A common approach is Zermelo–Fraenkel set theory ZF. It does not define a set as a "collection of objects" or something like that (although that is a nice intuitive understanding), but gives a list of axioms which have to be satisfied by things called sets.

However, it turns out that some "collections" of objects which one can easily name, for example

  • the collection of all sets,

  • the collection of all compact metric spaces,

cannot be sets (if they were sets, we would get contradictions as Russell's paradox). Such collections are called proper classes, and in a sense they are too big to satisfy all axioms of ZF. For example, there is no set containing as elements all sets and no set of all compact metric spaces.

Concerning your questions:

  1. The compact metric spaces do not form a set. This is not a drama, because it is possible to find a set of representatives. If you are satisfied with compact metrizable toplogical spaces (i.e. if you are not interested in specific metrics because all metrics generating the same compact topology on a given set are uniformly equivalent), then Daniel Wainfleet's comment shows what to do: The closed subspaces of the Hilbert cube $Q$ form a set $\mathfrak C$. All $X \in \mathfrak C$ are compact metrizable and each compact metrizable space is homeomorphic to some member of $\mathfrak C$. If you want to consider compact metric spaces, then it is just a little more complicated. The previous argument shows that each compact metric space $(X,d)$ admits a bijection $\phi : X \to X'$ to a subset $X'$ of $Q$. This $\phi$ induces a metric $d_\phi$ on $X'$ making $(X,d)$ and $(X',d_\phi)$ isometrically isomorphic. The metrics on a given set $X'$ form a set, thus 2. shows that all pairs $(X',d')$ where $X'$ is a subset of the Hilbert cube (i.e. a member of the power set of $Q$) and $d'$ is a metric on $X'$ also form a set $\mathfrak M$. Not all such $(X',d')$ are compact, but the subset $\mathfrak M_c$ of compact $(X',d') \in \mathfrak M$ gives you the desired set of representatives for the class of compact metric spaces.

  2. Yes. As Berci comments, this is the replacement axiom of ZF.

  • $\begingroup$ Thank you very much $\endgroup$ Nov 29, 2020 at 9:04

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